题目内容
如图所示,AB是某电场中的一条电场线.若有一电子以某一初速度,仅在电场力的作用下,沿AB由A运动到B,其速度图象如下图所示,下列关于A、B两点的电场强度EA、EB和电势
的判断正确的是:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124032664428.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241240327421925.png)
A.EA>EB B.EA<EB C.
D.![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124032789530.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124032508519.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124032664428.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241240327421925.png)
A.EA>EB B.EA<EB C.
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124032758546.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124032789530.png)
AC
分析:由图可知带电粒子速度变化情况,则可明确粒子在两点的加速度大小关系,即可确定电场强度的大小;
由功能关系可以确定电势的高低.
解答:解:由图可知,电子在A点加速度较大,则可知A点所受电场力较大,由F=Eq可知,A点的场强要大于B点场强;故A正确,B错误;
而电子从A到B的过程中,速度减小,动能减小,则可知电场力做负功,故电势能增加,故?
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124032805253.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124032836269.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124032852442.png)
故选AC.
点评:本题根据图象考查对电场的认识,要求学生能从图象中找出加速度的大小及速度的变化,再应用动能定理及牛顿第二定律进行分析判断;
同时还需注意,电势能是由电荷及电场共同决定的,故不能忽视了电荷的极性.
![](http://thumb2018.1010pic.com/images/loading.gif)
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